Complex Analysis - simply connected domain

• Nov 19th 2011, 06:03 AM
sinichko
Complex Analysis - simply connected domain

Prove that if domain $\displaystyle \Omega$ conformaly maps to unit disc $\displaystyle D=\left \{ \left | z \right |<1 \right \}$then $\displaystyle \Omega$ it's simply connected domain.

Thanks!
• Nov 19th 2011, 07:42 AM
Opalg
Re: Complex Analysis - simply connected domain
Quote:

Originally Posted by sinichko
Prove that if domain $\displaystyle \Omega$ conformaly maps to unit disc $\displaystyle D=\left \{ \left | z \right |<1 \right \}$then $\displaystyle \Omega$ it's simply connected domain.
You need to assume that the conformal map $\displaystyle f:\Omega\to D$ is bijective (and therefore invertible), otherwise the result is not true.
Suppose that S is a loop in $\displaystyle \Omega$. The image f(S) of S is a loop in the simply-connected space D. So there is a continuous transformation in D that contracts this loop to a point. The image in of that deformation under the map $\displaystyle f^{-1}$ will be a continuous transformation in $\displaystyle \Omega$ which contracts S to a point S to a point.
[QUOTE=Opalg;696240]You need to assume that the conformal map $\displaystyle f:\Omega\to D$ is bijective (and therefore invertible), otherwise the result is not true.